248
8 W IP Limiting Control Strategies
Table 8.1 Mean cycle time results (in hours) for Example 8.2 Iteration w=1 w=2 w=3 w=4 w=5
CT1(w) 0.200 0.235 0.260 0.277 0.290
CT2(w) 0.500 0.717 0.955 1.208 1.477
CT3(w) 0.500 0.696 0.897 1.099 1.299
∑ r j CT j (w) 1.150 1.578 2.021 2.475 2.936
Table 8.2 Workstation characteristics for Example 8.2 at a CONWIP level of 5 Measure CTk (5) λk (5) W IPk (5) uk (5)
Workstation 1 0.290 hr 1.703/hr 0.493 0.341
Workstation 2 1.477 hr 1.703/hr 2.515 0.852
Workstation 3 1.299 hr 1.533/hr 1.992 0.767
Notice that each utilization factors is less than 1; this will always be the case because the service rate serves as the upper limit to the throughput for each workstation, and the utilization factor equals the arrival rate (throughput rate) times the service time. More generally, the utilization factor is the arrival rate times the mean service time divided by the number of machines at the workstation; that is, uk (w) = λk (w) E[Ts(k)]/ck , where ck is the number of machines at Workstation k. Since 90% of the output from Workstation 3 is considered good, the throughput rate of this factory is 90% of the throughput rate of Workstation 3; thus, the mean number of jobs shipped from this factory is ths = 0.9 × 1.533 = 1.38/hr . Since the WIP level is always 5 for this example, the system mean cycle time is given by wmax 5 = = 3.62 hr . CTs = ths 1.38 • Suggestion: Do Problems 8.3, 8.4, 8.5 (a,b,c), 8.6 (a,b), and 8.7 (a,b). 8.1.1.2 Multi-Server Systems The algorithm for the multi-server case will require evaluation of the marginal probabilities associated with each workstation and is called a Marginal Distribution Analysis. (The probabilities are marginal in that they refer to the number of jobs within a specific workstation and not the joint probability of the number of jobs in different workstations at the same time.) As long as the processing times at each workstation are exponential, the analysis will yield the correct mean values. It is